A uniform circular disc has radius `R` and mass `m`. A particle also of mass `m` is fixed at a point `A` on the wedge of the disc as in fig. The disc can rotate freely about a fixed horizontal chord `PQ` that is at a distance `R//4` from the centre `C` of the disc. The line `AC` is perpendicular to `PQ`. Initially the disc is held vertical with the point `A` at its highest position. It is then allowed to fall so that it tarts rotating about `PQ`. Find the linear speed of the particle at it reaches its lowest position.
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Explanation

Let the angular velocity of ring be `omega`, when a particle is at the lowest position.
Taking `PQ` as reference level, appliying mechanical energy conservation between two positions
`mg(R+(R)/(4))+mg(R/4)= -mg(R+(R)/(4))-mg(R/4)+(1)/(2)I_(PQ)omega^(2)`
`3mgR=(1)/(2)I_(PQ)omega^(2)`
`I_(PQ)=m(R+(R)/(4))^(2)+[(1)/(4)mR^(2)+m((R)/(4))^(2)]=(15)/(8)mR^(2)`
`3mgR=(1)/(2)xx(15)/(8)mR^(2)omega^(2)implies omega=sqrt((16g)/(5R))`
Linear speed of particle is
`v=(R+(R)/(4))omega=(5R)/(4)sqrt((16g)/(5R))`
`=sqrt(5gR)`